Definition 11.6.1 (Lebesgue-Stieltjes Measure). Let $[a, b] \subset \real$, $E$ be a Banach space, and $G \in BV([a, b]; E^{*})$, then there exists a unique $\mu_{G} \in M_{R}([a, b]; E^{*})$ such that:
For any $f \in \text{Reg}([a, b], G; E)$,
\[\int_{a}^{b} f(t) G(dt) = \int_{[a, b]}\dpn{f(t), \mu_G(dt)}{E}\]For any $a \le c < d < b$,
\[\mu_{G}((c, d]) = G(d+) - G(c+) = \lim_{z \downto d}G(z) - \lim_{z \downto c}G(z)\]
The measure $\mu_{G}$ is the Lebesgue-Stieltjes measure associated with $G$.
Proof. By Proposition 11.4.3, the mapping $f \mapsto \int f(t)G(dt)$ is a continuous linear functional on $C_{0}([a, b]; E) = C([a, b]; E)$. By Singer’s Representation Theorem, there exists $\mu_{G} \in M_{R}([a, b]; E^{*})$ such that (1) holds for $C([a, b], G; E)$.
(2): Fix $x \in [a, b)$. For each $u \in (x, b)$, there exists a non-increasing function $\phi_{u} \in C_{c}([a, b))$ such that $\phi_{u}|_{[a, x]}= 1$ and $\supp{\phi_u}\subset [a, u]$. By outer regularity of $\mu_{G}$,
For each $u > x$, using integration by parts,
Let $v \in (x, u)$ and $(P = \bracs{x_j}_{0}^{n}, c = \bracs{c_j}_{1}^{n})$ be a tagged partition containing $\bracs{x, v, u}$, then since $\phi_{u}$ is non-increasing,
Since the above holds for all $v \in (x, u)$, $\int_{a}^{b} Gd\phi_{u} \in -\overline{G((a, u])}$, and
As $E^{*}$ is a Banach space, $G(x+)$ exists by Proposition 11.2.4, so
(1): Let $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]}\in RS([a, b], G)$ with
By linearity, for any regulated step map $f: [a, b] \to E$, $\int_{a}^{b} f(t) G(dt) = \int_{[a, b]}\dpn{f(t), \mu_G(dt)}{E}$. Therefore (1) holds for $\text{Reg}([a, b], G; E)$ by the Linear Extension Theorem.$\square$